Problem: The polynomial $P$ is graphed. ${1}$ ${2}$ ${3}$ ${\llap{-}1}$ ${\llap{-}2}$ ${\llap{-}3}$ ${1}$ ${2}$ ${3}$ ${4}$ ${\llap{-}1}$ ${\llap{-}2}$ $y$ $x$ $ P$ Graph of cubic polynomial P that passes through the points: negative 2, 1; -1, 3; 1, 0; and 2, 4. What is the remainder when $P(x)$ is divided by $(x+1)$ ?
Explanation: We can use the polynomial remainder theorem to solve this problem: For a polynomial $p(x)$ and a number $a$, the remainder on division by $x-a$ is $p(a)$. According to the theorem, the remainder when $P(x)$ is divided by $(x+1)$, which can be written as $(x-({-1}))$, is $P({-1})$. According to the graph, $P({-1})=3$. ${1}$ ${2}$ ${3}$ ${\llap{-}1}$ ${\llap{-}2}$ ${\llap{-}3}$ ${1}$ ${2}$ ${3}$ ${4}$ ${\llap{-}1}$ ${\llap{-}2}$ $y$ $x$ $ P$ Same graph as above with point marked at negative 1, 3 In conclusion, the remainder when $P(x)$ is divided by $(x+1)$ is $3$.